AFMC AFMC Solved Paper-2006

  • question_answer
    300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking \[g=10m/{{s}^{2}},\]work done against friction is:

    A) 200 J                                     

    B) 100 J

    C) zero                                      

    D) 1000 J

    Correct Answer: B

    Solution :

    Net work done in sliding a body up to a height h on inclined plane = Work done against gravitational force + Work done against frictional force \[\Rightarrow \]     \[W={{W}_{g}}+{{W}_{f}}\]                                  ?(i)                          but           W = 300 J \[{{W}_{g}}=mgh=2\times 10\times 10=20\,0\,J\] Putting in Eq. (i), we get \[300=200+{{W}_{f}}\] \[\Rightarrow \]     \[{{W}_{f}}=300-200\]            \[=100\,\,J\]


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